Problem: Find $\lim_{x\to\infty} \left(1+\dfrac{2}{x}\right)^{3x}$. Choose 1 answer: Choose 1 answer: (Choice A) A $e^3$ (Choice B) B $e^6$ (Choice C) C $\dfrac18$ (Choice D) D The limit doesn't exist.
Explanation: Taking $x$ to $\infty$ in $\left(1+\dfrac{2}{x}\right)^{3x}$ results in the indeterminate form $1^{^{\infty}}$. To make the expression easier to analyze, let's take its natural log (this is a common trick when dealing with composite exponential functions). In other words, letting $y=\left(1+\dfrac{2}{x}\right)^{3x}$, we will find $\lim_{x\to \infty}\ln(y)$. Once we find it, we will be able to find $\lim_{x\to \infty}y$. $\ln(y) =\dfrac{3\ln\left(1+\dfrac{2}{x}\right)}{x^{-1}}$ Taking $x$ to $\infty$ in $\dfrac{3\ln\left(1+\dfrac{2}{x}\right)}{x^{-1}}$ results in the indeterminate form $\dfrac{0}{0}$, so now it's L'Hôpital's rule's turn to help us with our quest! $\begin{aligned} &\phantom{=}\lim_{x\to \infty}\ln(y) \\\\ &=\lim_{x\to \infty}\dfrac{3\ln\left(1+\dfrac{2}{x}\right)}{x^{-1}} \\\\ &=\lim_{x\to \infty}\dfrac{\dfrac{d}{dx}\left[3\ln\left(1+\dfrac{2}{x}\right)\right]}{\dfrac{d}{dx}[x^{-1}]} \gray{\text{L'Hôpital's rule}} \\\\ &=\lim_{x\to \infty}\dfrac{\left[\dfrac{-6}{x^2\left(1+\dfrac{2}{x}\right)}\right]}{-x^{-2}} \\\\ &=\lim_{x\to \infty}{\left[\dfrac{-6}{-x^{-2}\cdot x^2\left(1+\dfrac{2}{x}\right)}\right]} \\\\ &=\lim_{x\to \infty}{\left[\dfrac{-6}{-\left(1+\dfrac{2}{x}\right)}\right]} \\\\ &=\dfrac{-6}{-1} \\\\ &=6 \end{aligned}$ Note that we were only able to use L'Hôpital's rule because the limit $\lim_{x\to \infty}\dfrac{\dfrac{d}{dx}\left[3\ln\left(1+\dfrac{2}{x}\right)\right]}{\dfrac{d}{dx}[x^{-1}]}$ actually exists. We found that $\lim_{x\to \infty}\ln(y)=6$, which means $\lim_{x\to \infty}y=e^6$. [Why?] In conclusion, $\lim_{x\to \infty}\left(1+\dfrac{2}{x}\right)^{3x}=e^6$.